Question

.11 0.0166 ().- . +x A 0.125- 0.010-X 0.11+x 0.115 0 PKa--log (1.8x10-4)= 3.74 0.12 PH: 3.74 +loglo. 1210.415) - 3.16 20. What would be the pH during a titration if 25.0 mL of 0.44 M acetic acid (HC2H302; K, = 1.8x10 ) were titrated with 50.0 mL of 0.22 M NaOH? (10 pts) A. 5.04 B. 5.61 C. 7.00 D. 8.39 E. 8.96 (7 pts... pOH) (4 pts... used moles in ICE table, and pOH) (0 pts... thinking all equivalence points are neutral pH) (7 pts... used moles in ICE table) (10 pts) HC₂H₂O2lag totliag) > C2H30zlags + H₂0 (1)

Answer 1

mol x0.44 1 L 20) No of moles of CH COOH = 25.0 mLx- 0.011 mol 1000 mL 1 L mol x0.22 50.0 mLx 0.011 mol No of moles of NaOH added 1000 mL L At equival ence point, pH is determined by the hy drolysis of the conjugate base of the weak acid: No of moles of CH COO no ofmoles of CH4COOH 0.011 mol Total volume (25.0+50.0) mL 75.0 mL 0.0750 L 0.011 mol Concentrati on of CH,COO", CHCOO 0.15 M = 0.0750 L ICE table CH2COO H2O CH, СООН + он I(M) с (М): E (M) Acidionization constant of CH C0OH, 0.15 0 0 x x - x 0.15-x 1.8x10 1.0x10- 1.8x103 К, Baseionization constant of CH COO,K 5.6x10-10 [сн,соон][он- CH,Cоо-] (x)(x) (0.15-x) x 9.1x10 M Baseionization constant 5.6x10-10 According to the equilibrium table, OH1= (x) M = 9.1x10 M [5O]O 1.0x10-4 According to ionic product of water, H2O 9.1x10- |он 1.1x10M -log H20*-1og (1.1x10) = 8.96 pH